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Category: Arithmetic

Question-146157

Question Number 146157 by henderson last updated on 11/Jul/21 Answered by gsk2684 last updated on 11/Jul/21 $$\underset{\frac{\mathrm{1}}{{e}}} {\overset{\lambda} {\int}}\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{ln}\:{x}\right)\frac{{d}\left(\mathrm{1}−\mathrm{ln}\:{x}\right)}{−\mathrm{1}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\left(\mathrm{1}−\mathrm{ln}\:{x}\right)^{\mathrm{2}} }{\mathrm{2}}\right]_{\frac{\mathrm{1}}{{e}}} ^{\lambda} \\ $$$$−\frac{\mathrm{1}}{\mathrm{4}}\left[\left(\mathrm{1}−\mathrm{ln}\:\lambda\right)^{\mathrm{2}}…

2-k-1-3-k-2-0-mod-k-5-min-k-k-N-

Question Number 146130 by Huy last updated on 11/Jul/21 $$\:\:\left(\mathrm{2}^{\mathrm{k}} +\mathrm{1}\right)\left(\mathrm{3}^{\mathrm{k}} +\mathrm{2}\right)\equiv\mathrm{0}\left(\mathrm{mod}\:\mathrm{k}+\mathrm{5}\right) \\ $$$$\:\mathrm{min}\:\mathrm{k}=?\:\:\:\left(\mathrm{k}\in\mathbb{N}\right) \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com

n-0-1-81-n-3n-3-

Question Number 145982 by ArielVyny last updated on 10/Jul/21 $$\underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\frac{\mathrm{1}}{\mathrm{81}}\right)^{{n}} \Gamma\left(\mathrm{3}{n}+\mathrm{3}\right)=?? \\ $$ Answered by Olaf_Thorendsen last updated on 10/Jul/21 $$\mathrm{diverges} \\ $$ Terms…

Question-80230

Question Number 80230 by M±th+et£s last updated on 01/Feb/20 Answered by mind is power last updated on 03/Feb/20 $${let}\:{f}\left({z}\right)=\frac{\pi{cot}\left(\pi{z}\right)}{\mathrm{1}+{z}^{\mathrm{4}} },{Main}\:{idee}\:{use}\:{Residus}\:{Th} \\ $$$${and}\:{evaluate}\:{Integrales}\:{in}\:{Two}\:{ways} \\ $$$${pols}\:{of}\:{f}\:{are}\:\:\:{n}\in\mathbb{Z}\cup\left\{{e}^{{i}\left(\frac{\mathrm{1}+\mathrm{2}{k}}{\mathrm{4}}\right\}\pi} ,{k}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}\right\}…

Question-80207

Question Number 80207 by peter frank last updated on 01/Feb/20 Commented by john santu last updated on 01/Feb/20 $$\left(\mathrm{2}\right)\:{area}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mid\begin{vmatrix}{\mathrm{2}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{3}\:\:−\mathrm{2}}\end{vmatrix}+\begin{vmatrix}{\mathrm{3}\:\:\:\:\:\:−\mathrm{2}}\\{{x}\:\:\:\:{x}+\mathrm{3}}\end{vmatrix}+\begin{vmatrix}{{x}\:\:\:\:{x}+\mathrm{3}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\mathrm{1}}\end{vmatrix}\mid \\ $$$$\Rightarrow\:\mathrm{5}\:=\frac{\mathrm{1}}{\mathrm{2}}\mid\left(−\mathrm{7}\right)+\left(\mathrm{5}{x}+\mathrm{9}\right)+\left(−{x}−\mathrm{6}\right)\mid \\ $$$$\Rightarrow\:\mathrm{10}\:=\:\mid\mathrm{4}{x}−\mathrm{4}\mid\: \\ $$$$\pm\:\mathrm{5}\:=\:\mathrm{2}{x}−\mathrm{2}\:,\:\mathrm{2}{x}=\mathrm{2}\pm\mathrm{5}\:\Rightarrow\begin{cases}{{x}=\frac{\mathrm{7}}{\mathrm{3}},\:{y}=\frac{\mathrm{16}}{\mathrm{3}}}\\{{x}=−\frac{\mathrm{3}}{\mathrm{2}},\:{y}=\frac{\mathrm{3}}{\mathrm{2}}}\end{cases}…